Question

A two-digit number is such that the product of the digits is 16. When 54 is subtracted from the number, the digits are interchanged. Find the number.

Answer 1

Hint: While solving a question, which involves finding a two-digit number with given properties, we start by assuming the number as xy. Further, this can be written as $(10\times x+y)$. To explain, say we write a two-digit number (say 12), we can write
12 = $10\times 1+2$
After representing $(10\times x+y)$, we can use the given properties now to find x and y, to get the two-digit number.

Complete step-by-step answer:

Firstly, it is given that the product of the digits is 16. (The digits are x and y) So, we have,
xy = 16 - (1)
The second condition given to us is that when 54 is subtracted from the number, the digits are interchanged. Thus, we would have,
\[10\times x\text{ }+\text{ }y\text{ }\text{ }-54\text{ }=\text{ }10\times y\text{ }+\text{ }x\]- (2)
To explain expression (2), we express the number xy as $10\times x+y$ (as explained in the hint). Also, the reverse of the number would be yx, which can be represented as $10\times y+x$.
Now, we solve the expression (2)
$\begin{align}
  & 10x-x-54+y=10y \\
 & 10x-x-54=10y-y \\
\end{align}$
9x - 54 = 9y
9(x - y) = 54
x - y = 6 - (3)
Now we solve (1) and (3)
From (1), we have,
(x)y = 16
y = $\dfrac{16}{x}$
We put this value of y in (1), we get,
$\begin{align}
  & x-\dfrac{16}{x}=6 \\
 & \left( x-\frac{16}{x} \right)\times x=6\times x \\
 & {{x}^{2}}-16=6x \\
 & {{x}^{2}}-6x-16=0 \\
 & {{x}^{2}}-(8x-2x)-16=0 \\
 & {{x}^{2}}-8x+2x-16=0 \\
 & x(x-8)+2(x-8)=0 \\
 & (x+2)(x-8)=0 \\
\end{align}$
Thus, we get x = -2 or x = 8
We remove x = -2, since, x is a positive number.
Thus, x = 8.
Now, since, xy = 16,
So,
 $\begin{align}
  & y=\frac{16}{x} \\
 & y=\frac{16}{8} \\
 & y=2 \\
\end{align}$
Thus, the required number, xy, is 82.

Note: An alternative to solve the problem is to put the value of x from (1) We had, x – y =6, thus, x = y+6.
Now we can put this in (3). Since, we put in xy = 16, thus, we would get,
$\begin{align}
  & (y+6)\times y=16 \\
 & {{y}^{2}}+6y-16=0 \\
 & {{y}^{2}}+8y-2y-16=0 \\
 & y(y+8)-2(y+8)=0 \\
 & (y-2)(y+8)=0 \\
\end{align}$
Thus, we get, y=2 or y=-8. Now, we can remove, y=-8, since, y should be positive.
Thus, we get, y=2. Now, since, xy=16
We get, x=8. Thus, once again, we get 82 as the two digit number.